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Fluid Mechanics - Engineering

Q1:

A Newtonion liquid (ρ = density, μ = viscosity) is flowing with velocity v in a tube of diameter 'D'. Let Δp be the pressure drop across the length 'L'. For a laminar flow, Δp is proportional to

A L ρv2/D

B L μ;V/D2

C D ρv2/L

D μ V/L

ANS:A - L ρv2/D

For laminar flow of a Newtonian fluid through a tube of diameter DDD and length LLL, the pressure drop Δp\Delta pΔp is proportional to: Δp∝LμvD2\Delta p \propto \frac{L \mu v}{D^2}Δp∝D2Lμv​ Here's an explanation of each term:

  • LLL: Length of the tube.
  • μ\muμ: Dynamic viscosity of the fluid.
  • vvv: Average velocity of the fluid.
  • DDD: Diameter of the tube.
Therefore, the correct proportional relationship for the pressure drop Δp\Delta pΔp in laminar flow is: Δp∝LμvD2\Delta p \propto \frac{L \mu v}{D^2}Δp∝D2Lμv​ This indicates that the pressure drop is directly proportional to the length LLL, the dynamic viscosity μ\muμ, and the average velocity vvv, and inversely proportional to the square of the diameter DDD.